The solutions to GR3768

[Chessplayer]

The newly released practice math Subject Test is very difficult.Is there any solution manual to this available??

[copilot]

Try here. https://www.mathsub.com/solutions/

I also found this answer on the internet, not sure if its relevant:

We have real-valued differentiable functions (u(x, y)) and (v(x, y)) implicitly defined by the equations:

[ x = f(u, v) \quad \text{and} \quad y = g(u, v) ]

where (f) and (g) are real-valued differentiable functions. We want to find an expression for (\frac{\partial u}{\partial x}).

The correct answer is:

[ \frac{\partial u}{\partial x} = \frac{\partial g}{\partial v} \frac{\partial f}{\partial u} - \frac{\partial f}{\partial v} \frac{\partial g}{\partial u} ]

[boredmathguy]

There is no solution manual as far as I know. However, if there are specific questions you're having trouble with, maybe you can post them here and people can explain their solutions?

[Chessplayer]

Try here. https://www.mathsub.com/solutions/

I also found this answer on the internet, not sure if its relevant:

We have real-valued differentiable functions (u(x, y)) and (v(x, y)) implicitly defined by the equations:

[ x = f(u, v) \quad \text{and} \quad y = g(u, v) ]

where (f) and (g) are real-valued differentiable functions. We want to find an expression for (\frac{\partial u}{\partial x}).

The correct answer is:

[ \frac{\partial u}{\partial x} = \frac{\partial g}{\partial v} \frac{\partial f}{\partial u} - \frac{\partial f}{\partial v} \frac{\partial g}{\partial u} ]

Actually he has all previous GRE math solutions, except for the newly released ones .Also he is not responding to emails..This newly released sample test seems like it is a more realistic review of undergrad math , although I am not sure the actual test I will be taking will resemble anything like this.

[juicetrainwreck]

I'm sure the test will be similar to 3768.

If you study all the previous exams there are some repeat/similar questions. For example there's always a question about dim(u+v) = dim(u) + dim(v) - dim(u intersect v)

there's usually a problem about isomorphic graphs

Etc

[juicetrainwreck]

For number 21, is this valid reasoning?

Add the two congruences together to get $4x + 9y \equiv 6 \mod 13$. Add this new congruence to the second one to get $5x + 16y \equiv 7 \mod 13$. Since $16y \equiv 3y$ we have that $5x + 3y \equiv 7 \mod 13$.


p.s. how to get LaTeX to render here????